Show that an ideal is proper if and only if it does not contain 1.

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- Jul 27th 2009, 01:11 PMpeteryellowring
Show that an ideal is proper if and only if it does not contain 1.

- Jul 27th 2009, 03:01 PMsiclar
Prove these two statements: If 1 is in the ideal, it is not proper. If 1 is not in the ideal, it is proper. The second statement is obvious by the definition of being a proper ideal, i.e. if the ring is $\displaystyle R$ and ideal is $\displaystyle I$, $\displaystyle I\neq R$ since $\displaystyle 1\in R, 1 \notin I$.

Use the definition of an (left) ideal: for all elements $\displaystyle r$ of the ring, $\displaystyle rI\subseteq I$. In particular, if $\displaystyle i\in I$ then $\displaystyle ri\in I$ for any $\displaystyle r$. What if $\displaystyle i=1$? - Jul 27th 2009, 09:07 PMGamma
Does your ring definition require all rings to be unitary?

If you take the ring to be $\displaystyle 2\mathbb{Z}$. Certainly $\displaystyle 2\mathbb{Z}\subset 2\mathbb{Z}$ and it does not contain 1, but is not a proper ideal since it is the whole ring.